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Proof step 108 from proof abs-paths-eq (previous step)

tree[p3 + p2].dir?, tree[p3].dir?, abs(tree, ino0)(dirs), ispath(p3 + p2, ino0, ino1, dirs, files), ispath(p3, ino0, ino1, dirs, files), ino1 ∈ dirs, p3 ∈ tree, p2 ∈ tree[p3], ino0 ∈ dirs, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ p2 = ε

User applied rule insert proof lemma

And obtained the following new goals

Rule argument

path: 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1 subst: [tree, p3, p, ino0, dirs, ino1, files] → [tree, p3, p2, ino0, dirs, ino1, files]

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